4.4 Parabola (e = 1)
A parabola is defined to be the set of points P in the plane whose distances from a fixed point F are equal to their distances from a fixed line d. We obtain a parabola in standard form if
the focus F lies on the xaxis, and has coordinates (a, 0), where a > 0;
the directrix d is the line with equation x = −a.
Thus the origin lies on the parabola, since it is equidistant from F and d.
Let P (x, y) be an arbitrary point on the parabola, and let M be the foot of the perpendicular from P to the directrix. Since FP = PM, by the definition of the parabola, it follows that FP^{2} = PM^{2}; that is,
(x − a)^{2} + (y − 0)^{2} = (x + a)^{2}.
Multiplying out the brackets, we obtain x^{2} − 2ax + a^{2} + y^{2} = x^{2} + 2ax + a^{2}, which simplifies to the equation y^{2} = 4ax.
Each point with coordinates of the form (at^{2}, 2at), where t ∈ , lies on the parabola, since (2at)^{2} = 4a × at^{2}. Conversely, we can write the coordinates of each point on the parabola in the form (at^{2}, 2at). If we choose t = y/(2a), then y = 2at and
as required. Thus there is a oneone correspondence between the real numbers t and the points of the parabola.
We call the xaxis the axis of a parabola in standard form, since the parabola is symmetric with respect to this line, and we call the origin the vertex of a parabola in standard form, since it is the point of intersection of the axis with the parabola. A parabola has no centre.
We summarise these facts as follows:
Parabola in standard form
A parabola in standard form has equation
It can also be described by the parametric equations
It has focus (a, 0) and directrix x = −a; its axis is the xaxis and its vertex is the origin.
Example 8
Consider the parabola with equation y^{2} = 2x and parametric equations , y = t (t ∈ ).
(a) Write down the focus, vertex, axis and directrix of the parabola.
(b) Determine the equation of the chord that joins the two distinct points P and Q with parameters t_{1} and t_{2}, respectively. Determine the condition on t_{1} and t_{2} such that the chord PQ passes through the focus of the parabola. Such a chord is called a focal chord.
Answer
(a) The parabola is in standard form, where 4a = 2, that is, . It follows that its focus is , its vertex is (0, 0), its axis is the xaxis and the equation of its directrix is .
(b) The coordinates of P and Q are and , respectively.
We must consider the cases and separately.
We suppose first that . Then the gradient of PQ is
Since lies on the line PQ, the equation of PQ is
Multiplying both sides by t_{1} + t_{2}, we obtain
so
and hence

If , then t_{1} = −t_{2} (as t_{1} ≠ t_{2}, since P and Q are distinct). When t_{1} = −t_{2}, PQ is parallel to the yaxis and the equation of PQ is
that is, 0 = 2x + t^{1}t^{2}.

Thus equation (4.5) is the equation of PQ in the case t_{1} = −t_{2} also, for then t_{1} + t_{2} = 0.
Thus PQ passes through the focus if and only if satisfies equation (4.5), which is the case if and only if
(t_{1} + t_{2})0 = 1 + t_{1}t_{2};
that is, if and only if t_{1}t_{2} = −1.
Example 55
Consider the parabola with equation y^{2} = x and parametric equations .
(a) Write down the focus, vertex, axis and directrix.
(b) Determine the equation of the chord that joins the two distinct points P and Q with parameters t_{1} and t_{2}, respectively.
(c) Determine the condition on t_{1} and t_{2}, and so on P and Q, such that the focus of the parabola is the midpoint of the chord PQ.
Answer
(a) The parabola is in standard form, where 4a = 1, that is, . It follows that the focus is , the vertex is (0, 0), the axis is the xaxis, and the equation of the directrix is .
(b) The coordinates of P and Q are and , respectively. It follows that if , then the gradient of PQ is
Since lies on the line PQ, it follows that the equation of PQ is
Multiplying both sides by 2(t_{1} + t_{2}), we obtain
so
and hence
If , then t_{1} = −t_{2} (as t_{1} ≠ −t_{2}, since P and Q are distinct). If t_{1} = −t_{2}, then PQ is parallel to the yaxis and the equation of PQ is
that is,
Thus equation (S.18) is the equation of PQ in the case t_{1} = −t_{2} also, for then t_{1} + t_{2} = 0.
(c) The midpoint of PQ is
This is the focus if and only if
which is the case if and only if t_{2} = −t_{1} and , that is, and so t_{1} = ± 1. Thus the midpoint of PQ is the focus if and only if P and Q are the points and .
Remark: The focus is the midpoint of PQ if and only if PQ is the chord through the focus parallel to the yaxis.